Solution:

The total average kinetic energy \( E \) of translatory motion for an ideal gas is given by:

\[
E = \frac{3}{2} nRT
\]

where:
- \( n \) is the number of moles,
- \( R \) is the gas constant (\( 8.314 \, \text{J/mol·K} \)),
- \( T \) is the temperature in Kelvin.

We can rearrange to find \( nRT \):

\[
nRT = \frac{2}{3} E
\]

The ideal gas law also gives us:

\[
PV = nRT
\]

Thus,

\[
P = \frac{nRT}{V} = \frac{\frac{2}{3} E}{V}
\]

Substitute \( E = 1.5 \times 10^5 \, \text{J} \) and \( V = 20 \, \text{litres} = 20 \times 10^{-3} \, \text{m}^3 \):

\[
P = \frac{\frac{2}{3} \times 1.5 \times 10^5}{20 \times 10^{-3}}
\]

\[
P = \frac{10^5}{20 \times 10^{-3}} = 5 \times 10^6 \, \text{N/m}^2
\]

So, the pressure is \( 5 \times 10^6 \, \text{N/m}^2 \).